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Re-Inventing The Wheel-Part1-Clemente_Figuera

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  • bistander
    replied
    Electromagnet force

    Originally posted by hanon1492 View Post
    The way to calculate it is with the equation: Power = Force • Velocity

    Taking as velocity the movement of the magnetic lines back and forth cutting the induced coil wires. Yo can go into further detail in this old post link. I recalculated that value of 14.8 lb/KW and I think it is adjusted for an induced coil length of 2 inches and a frequency of 60 Hz. If you are good with english units please confirm if I did it well or not. Thanks
    Hi hanon,

    Force does not relate power directly like you state. But I did find an article reducing certain properties to a generalized force statement as follows:

    The maximum force per unit area (which is pressure) which an iron core electromagnet can exert is approximately 145 lbf/in². See the physics section here: https://en.m.wikipedia.org/wiki/Electromagnet

    You had given a brief discussion on electromagnet force in post #1232 and included this from one of the references:



    Again you can see that the expression does not contain power or terms which could lead directly to a power or energy calculation. To do that, certain constraints must be defined.

    Originally posted by marathonman
    That 14.8 lbs force was given to me by meentor's figures but have lost contact to get all of equations. it had to do with horse power to watt conversion as explained to me. other conversions are necessary but i do not have them. all i have right now is the final outcome...... 14.8 lbs force per kilowatt. he was very specific about this.
    Perhaps MM could check with his source. Maybe I'm missing something and sure would like to know if that is the case.

    Regards,

    bi

    Leave a comment:


  • Elcheapo
    replied
    cores

    MM,

    Thanks for the helpful info.

    So I guess your large square 1.5x1.5x3" cores is what gets you that large 25 ohm reactance. Mine are only at 6 ohms reactance. So got a long ways to go to make improvement.
    Hope you get the new unit working in the near future.

    Leave a comment:


  • Elcheapo
    replied
    MM,

    "How in the world did you come up with that figure of 400 amps.??? that is impossible with two electromagnets and ressistive wire."

    Of course it's impossible. But you said ,
    " Like i have been telling everyone part G controls the currant not the primaries."

    You are implying that the primaries have no control over the current. Which is just as ridiculous as my 400 amps without the primaries.

    Of course once everything is set up, then it's part G that controls current, but coil reactance has a profound effect on the over-all current. no?

    All I'm asking from you is: Was the reactance of your primaries set at 25 ohms in order to get just 2 amps at 50 volts?
    Also,in your demo, were all 3 coils wound clockwise?

    Leave a comment:


  • hanon1492
    replied
    Originally posted by marathonman

    example; i will be using around 5 amp for high 2.5 amp for low at around 100 volts. this is a total of 750 watts plus 500 over is 1250. this 1250 is minimum needed to function properly. mine from above is 1566 va rating.

    MM
    If the toroid has to have room to storage all electromagnets wattage then perhaps all electromagnets in each set must be wired in parallel in order to join all their wattages in the toroid inlet, and from that point the thicker wire should be used. If wired in series then both the electromagnets and the toroid may use the same wire thickness if part G is switched with transistors because all them will see the same intensity. Just a thought

    From those numbers I see that the impedance of each set of electromagnet alone is 20 ohms (100volt/5 A) and adding the maximun impedance of the toroid+electromagnets is 40 ohms, which gives a impedance to the toroid of 20 ohms. Considering a low resistance system we can approximate that those are aprox. the inductive reactance of your part G, XL = 2•pi•f•L, then for 60 Hz your part G maximun inductance is 53 mH in each half turn. Just to have a reference.

    I have seen some online calculators to get the inductance of a toroid as function of its dimensions and number of turns. I do not know if those may be applied to a system with two opposite taps and two opposite fields.
    Last edited by hanon1492; 10-16-2016, 07:31 PM.

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  • Elcheapo
    replied
    Originally posted by marathonman
    Really,

    Like i have been telling everyone part G controls the currant not the primaries.

    MM
    MM,

    We all already know that part G controls the current. But that current still has to pass
    through the primaries which will also affect the current.

    Part G in your demo was nothing more than eight .125ohm resistors being switched in by commutator contacts.
    It takes no more than simple ohms law to calculate current with any known impedance.
    So how did you arrive at 2 amps & 50 volts using only part G?
    Either you used coils with 50 ohm XL or you just used part G to give you a current of 400 amps(50/.125)
    Wow thats a pretty big current. So which way was it?

    If you got a cop3 with your demo using resistors, then I should be able to get a cop4 or more using NO resistors.
    ps:
    I'll join up with the rest of the crowd when I can simulate your demo.

    Leave a comment:


  • Ufopolitics
    replied
    Originally posted by marathonman
    I understand, your just analyzing deeper then i chose to except for part G.
    Well, just having in mind that by fluctuating a virtual field along static iron and copper will induce a very strong output, higher than input... is enough to open a whole new horizon to many, many other possibilities, whether it was achieved with basic resistors...state of the art electronics or a part G...

    just so you know, my mentor used a 100 amp alternator core for part G so as i said it doesn't need to be huge.
    That is a great idea!!...I've got many of those...plus also AC Induction motors stators would be good as well.

    On a separate note...have you realized that your Toroid Part G is just doing the same exact operation that two primaries opposite at unison do every 180º?

    It is just that part G does not have a secondary within the core...but primaries are its "secondaries"...

    That is the way I see this...I may be wrong.


    Regards


    Ufopolitics
    Last edited by Ufopolitics; 10-16-2016, 04:26 PM.

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  • Ufopolitics
    replied
    Originally posted by marathonman
    Yes, i agree but certain guidelines must be followed and all i was doing was throwing them out there. weather you people choose to follow as was passed to me is entirely up to you.

    if you built part G with your guidelines it will end up so heavy you need a dolly to move it. there is no need for it to be so big and expensive just cover expended wattage plus headroom.

    happy building.

    MM
    MM,

    ...Is not that...I rather concentrate on the pure primary and expand-retract field into the secondary...you know, the basic principle.

    I consider "A Module" based on two primaries and one secondary in between...so, once I get that the total output on one module, it could be reproduced as your needs.

    It is just that I rather concentrate on The Magnetic Field at Primaries first.

    I am trying to build the most compact and less expensive primaries...so I don't need a dolly for part G Toroid...


    Take care



    Ufopolitics
    Last edited by Ufopolitics; 10-16-2016, 03:46 PM.

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  • Ufopolitics
    replied
    Originally posted by marathonman
    I'm not following you.

    First of all you must decide how much output you need and that will dictate the size of secondary core and coil. then build your primaries to sustain that output with your choice of voltage and amperage combination remembering each primary is accountable for half of the secondary output. then use your total wattage expended for primary operation to calculate size of part G plus headroom.

    example; i will be using around 5 amp for high 2.5 amp for low at around 100 volts. this is a total of 750 watts plus 500 over is 1250. this 1250 is minimum needed to function properly. mine from above is 1566 va rating.

    MM
    Is Ok...it is just we both have different ways to build this same device, that's all.




    Ufopolitics

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  • Ufopolitics
    replied
    Originally posted by marathonman

    UFOP;

    Total mass of Part G must match total wattage expended by hi and low primary sets with head room......ie. 500 va over. no need to over complicate.
    mass does not need to be equal to primaries just total wattage plus headroom.


    MM
    Hi MM,

    So that means you already have all your primaries set built up and they do the expected output at secondaries in order to build part G Toroid right?


    Regards


    Ufopolitics

    Leave a comment:


  • Ufopolitics
    replied
    The order I am working on Figuera...

    Hello to All,

    While I see how you all guys are developing the Figuera device...I wanted to share the order I am working on it.

    This device depends MAINLY on the atmospheric or SPATIAL FLUCTUATIONS of the magnetic fields at PRIMARIES, in order to function properly...So, my FIRST focusing is on Primaries construction, and even reduced to just ONE Primary electromagnet design, in order to observe its expansion and retraction. Before I proceed to reproduce more iron cores and windings...

    Once we have just one primary iron core size wound with low resistance wire, then, in order to check FULL FIELD EXPANSION, we must add the secondary core (without any windings whatsoever), and see how this secondary iron core expands our primary field without major decay.

    1-If we all resume the primaries operation, we realize it bolts down to a HIGH FIELD and a LOW FIELD basically.

    2-However, the TRANSITION between Hi-Lo Field must be done very smoothly...meaning NOT RADICAL, NOT ABRUPT, NOT A STEEP DROP OFF.

    But first we need number one....achieving a Hi and a Lo Field without collapsing field all the way.

    High Field is basically dependent upon the primary core and wire design...and we are not looking to spent hundred of watts on this endeavor, but exactly all we need to expand field along the secondary core length.

    Many do not believe in the screening of a CRT to see the magnetic field(s)...well, too bad because I do...and it helps me to visualize it perfectly and with so much accuracy as the signal we read in our scopes. Basically a simple Black and White old small TV, that I have disconnected its vertical coil at Cathode Tube...so it only shows me a horizontal line...is enough to observe Hi and Lo Field.

    Low Field...before spending a lot of money on a bunch of resistors...I used a Potentiometer and the wiper set at the extreme opposite end further away from the straight positive connection or High Field....then I test to reach lowest spot before my horizontal line becomes flat line (dead field)...just a bit above will simulate as Figuera writes..."Field moves further away"...So, once I got this perfect Lo Field, then I measure wiper total resistance and divide it by the number of segments at commutator...that gives me the total series resistors I would need to install, making the "FALL" from Hi to Lo in a smooth fashion. And here is understood that the more segments we have at commutator, the smoother the fluctuations would take place.

    Under a B&W CRT horizontal line screening We could see the field response and intensity under fluctuations of current changes...just like flashing a powerful bulb on a dark screen.

    The way I understand Part G Toroid...is that Primaries design, basically their iron core TOTAL mass (Which means and INCLUDES ALL Primaries involved in our set up), DICTATES the size of our Full Iron Toroid....

    So, basically...this is the reason why... I am still working on Primaries...and again, I am just sharing above the way I am proceeding with the development of my build here, just like "thinking out loud"...and by no means I want anyone to do or necessarily believe in what I have written above...


    Regards


    Ufopolitics
    Last edited by Ufopolitics; 10-16-2016, 01:56 PM.

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  • Ufopolitics
    replied
    Originally posted by seaad View Post
    ""and always in contact with two of their contacts""
    Seaad,

    I already answer to above...don't you believe me?

    It means the brush must be wider than one commutator segment, could be one and a half...and this is done in order NOT to allow Field to EVER Collapse, so like MM wrote..."Make BEFORE Brake"...Which means that if brush is the same width as contacting segment, when both are fully aligned/engaged then either previous or next would be off...NOT GOOD!

    Originally posted by seaad View Post
    ""without any more complications than the turning of a brush or group of brushes that move circularly around the cylinder “G” "" Pat. 44267
    Back in 1908... Semiconductors, specifically Diodes were not discovered yet...so, Figuera as Buforn were depending on a second commutator plus brushes in order to convert the secondaries output from AC to DC...did you know that was the way back then?...when diodes were still not available on the market?


    Hope you understand now.


    Ufopolitics
    Last edited by Ufopolitics; 10-16-2016, 01:09 PM.

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  • hanon1492
    replied
    Originally posted by marathonman
    Really,

    Like i have been telling everyone part G controls the currant not the primaries.

    MM
    I believe that was not the kind of answer Elcheapo was expecting after the time and effort he surely have dedicated to replicate that design. At least tell him what was the total value of your resistance wire , in ohms, in order to regulate the current.

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  • seaad
    replied
    TWO of their contacts

    ""and always in contact with two of their contacts""

    ""without any more complications than the turning of a brush or group of brushes that move circularly around the cylinder “G” "" Pat. 44267
    Attached Files

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  • Elcheapo
    replied
    demo

    MM,

    I need your help. Did a spread sheet with your data for the cop3 demo. Copy is attached.
    As you can see, if using 50 volts, the total current for both hi side & low side would be 12 amps.
    So 12 x 50 gives you 600 watts as the power input. A far cry from 100W.
    This is all dependent of course on the coil's reactance. My coils are wound with 340 turns of 19awg
    on iron cores of 1.5" dia. & 2.75" long. XL checks out at 6 ohms using 60 hz. Close to yours.
    Not knowing the XL of your coils, I used a value of 7 ohms.

    Take note of the chart that the total current for both hi & lo is always at 12 amps.
    So current never CHANGES in this 3 coil system. Only the B field should vary up & down.(no lenz)

    You said your input was 50 volts @ 2 amps. Did you give me some wrong info?
    If not then tell me what I have to change to make this thing work like you did.
    Attached Files

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  • bistander
    replied
    Comm jumpers

    The jumpers on the commutator just cut in half the number of wires needed between the commutator and resistor bank.

    Leave a comment:

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